One knows that in higher category theory, the category of $(\infty,n-1)$ categories is naturally an $(\infty,n)$ category ,(I use the word category to mean category in the correct weakened sense). When the category of $(\infty,1)$ categories is regarded as a weakened kan complex, we may regard these objects as a full subcategory of simplicial sets. This is a category in the strict sense. One ought to expect that associativity of the maps between the weakened kan complex be some sort of weakened associativity. The question is: is this weakened associativity there, and if so how is it understood?

1 Answer
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You can see the collection of $(\infty,1)$-categories as forming themselves an $(\infty,1)$-category, which is sufficient to see where weak associativity shows up: There are many models for the intuiti9ve concept of $(\infty,1)$-category, the simplest is that of a usual 1-category endowed with a class of weak equivalences (see Barwick/Kan's "Relative Categories: Another model for the homotopy theory of homotopy theories" here).

With the weak Kan complexes - together with the notion of equivalence between them - you happen to have found a strictly associative model for the $(\infty,1)$-category of $(\infty,1)$-categories. You can transform it into different other models, e.g. into simplicially enriched categories or quasicategories or Segal categories, as exposed e.g. in Bergner's "Three models for the homotopy theory of homotopy theories" (available here).

The Segal category and the quasicategory of $(\infty,1)$-cats do no longer have strict associativity and the fact that they are equivalent descriptions of the $(\infty,1)$-cat of $(\infty,1)$-cats reflects that the strict associativity in your model was an accident and not an essential feature...

Edit (in response to the comment) About the significance of having a strict model: Well, different models have different advantages. Your strict one is certainly good for computing compositions of functors between $(\infty,1)$-cats. The quasicategory of $(\infty,1)$-cats on the other hand is e.g. a better model to relate the $(\infty,1)$-cat of $(\infty,1)$-cats to other $(\infty,1)$-cats - examples are the relations between the quasicategories of (small) $(\infty,1)$-cats, presentable $(\infty,1)$-cats and stable $(\infty,1)$-cats given in Lurie's "Higher Topoi" and in DAG 1 (now "Higher Algebra"): There are $(\infty,1)$-adjunctions between these - e.g. between $(\infty,1)$-cats and stable $(\infty,1)$-cats given by taking spectra in an $(\infty,1)$-cat, forgetting the stability of a stable $(\infty,1)$-cat, respectively - and these facts would be hard to express using your model.

This is a nice answer. You went beyond answering my question. I wish that i could upvote this twice. One question that I have is on the significance of having a model with strict associativity. This seems to say that the theory itself can be simplified somewhat.
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Lunasaurus Rex Aug 27 '11 at 7:46

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It is not that significant in itself. One of the equivalent models of all $(\infty, 1)$-categories is strict categories enriched in simplicial spaces (even Kan complexes). So this means that EVERY $(\infty,1)$-category has a strictly associative model. (As Peter rightly notes though, computing with the strict model is difficult.)
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Chris Schommer-PriesAug 28 '11 at 14:39